What is the vertices f(x)=2x^2+4x-2 ?

The vertices f(x)=2x^2+4x-2 is Minimum (-1,-4)

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vertices f(x)=2x^2+4x-2

vertices

f (x) = 2 x^{2} + 4 x - 2

Minimum (- 1, - 4)

Solution steps

y = 2 x^{2} + 4 x - 2

The parabola parameters are:

a = 2, b = 4, c = - 2

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 \cdot 2}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(- 1, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, - 4)

f (x) = 2 x^{2} + 4 x - 3

y = x^{2} - 5 x 3

f (x) = 2 x^{2} + 9 x + 10

f (x) = x^{2} + 6 x - 16

x = 1 - y^{2}

What is the vertices f(x)=2x^2+4x-2 ?
The vertices f(x)=2x^2+4x-2 is Minimum (-1,-4)